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Author Topic: Equivalence relation แบ่งสมาชิกออกเป็น distinct classes  (Read 10444 times)
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เกียรติศักดิ์
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« on: November 17, 2006, 12:34:45 AM »
Equivalence relation แบ่งสมาชิกออกเป็น distinct classes.

Instead of the proof demonstrated in class*, can I also prove this statement by proving that

"If there are two set of equivalent elements, and one element from one set is equal to an element from the other set, then the two sets are considered the same class; that is, no different classes have the same element--the classes are distinct."?

* which prove that by proving that "If there are two classes, and one element from one class is equal to an element from the other class, then the two classes are equal.".

It is what I did in class and not sure whether it is a valid proof.
« Last Edit: November 17, 2006, 12:36:43 AM by เกียรติศักดิ์ » Logged
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« Reply #1 on: November 17, 2006, 06:04:05 AM »
Quote from: เกียรติศักดิ์ on November 17, 2006, 12:34:45 AM
...
* which prove that by proving that "If there are two classes, and one element from one class is equal to an element from the other class, then the two classes are equal.".
...

How do you prove that two classes are equal? Isn't it the same method that we did in the class?  coolsmiley
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« Reply #2 on: November 17, 2006, 10:23:53 AM »
Quote from: ปิยพงษ์ - Head Admin on November 17, 2006, 06:04:05 AM
Quote from: เกียรติศักดิ์ on November 17, 2006, 12:34:45 AM
...
* which prove that by proving that "If there are two classes, and one element from one class is equal to an element from the other class, then the two classes are equal.".
...

How do you prove that two classes are equal? Isn't it the same method that we did in the class?  coolsmiley

That's what we did in class on the blackboard (by starting with that claim; I gave it as a reference). I wonder whether it is correct to start with this alternative claim in the very beginning:
Quote from: เกียรติศักดิ์ on November 17, 2006, 12:34:45 AM
If there are two set of equivalent elements, and one element from one set is equal to an element from the other set, then the two sets are considered the same class; that is, no different classes have the same element--the classes are distinct."
« Last Edit: November 17, 2006, 11:06:27 AM by เกียรติศักดิ์ » Logged
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« Reply #3 on: November 17, 2006, 11:05:33 AM »
Here's how it went in my paper.

Equivalence relation แบ่งสมาชิกออกเป็น distinct classes.
Claim: If there are two sets of equivalent elements, and one element from one set is equal to an element from the other set, then the two sets are considered the same class; that is, no different classes have the same element--the classes are distinct.

Proof: Supposed that there are two sets of equivalent elements
\{g_1, g_2, \cdots, g_r\} and \{h_1, h_2, \cdots, h_s\} .

Let g_i = h_j , then g_i \equiv h_j (reflexivity).
But g_i \equiv \forall g , so h_j \equiv \forall g (transitivity),
and h_j \equiv \forall h , so g_i \equiv \forall h (transitivity).

Since each element of one set is equivalent to all elements of the other set,
the two sets can be merged into a single class.

Thus, two sets cannot be considered two classes if there exists at least one common element.
Therefore, two classes are distinct.
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